Mathematics — Standard (041) — CBSE Class 10 board question

Expanding the LHS:

$(\sin A + \cosec A)^2 + (\cos A + \sec A)^2$

$= \sin^2 A + 2\sin A \cdot \cosec A + \cosec^2 A + \cos^2 A + 2\cos A \cdot \sec A + \sec^2 A$

$= (\sin^2 A + \cos^2 A) + 2(1) + 2(1) + \cosec^2 A + \sec^2 A$

$= 1 + 4 + (1 + \cot^2 A) + (1 + \tan^2 A)$

$= 7 + \tan^2 A + \cot^2 A$

Comparing with $x + \tan^2 A + \cot^2 A$:

$$\boxed{x = 7}$$

Source: Exercise 8.3, Q.4(viii), Chapter 8

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• Expand both squares and use $\sin A \cdot \cosec A = 1$ and $\cos A \cdot \sec A = 1$.
• Apply $\sin^2 A + \cos^2 A = 1$, $\cosec^2 A = 1 + \cot^2 A$, and $\sec^2 A = 1 + \tan^2 A$.
• The constant terms add up to $1 + 2 + 2 + 1 + 1 = 7$, so $x = 7$.
• This identity is directly stated in Exercise 8.3 Q.4(viii) of the textbook.